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geerky42
 one year ago
Does Batman Equation really work?
And can we "union" relations?
(Sorry for long reply.)
geerky42
 one year ago
Does Batman Equation really work? And can we "union" relations? (Sorry for long reply.)

This Question is Closed

geerky42
 one year ago
Best ResponseYou've already chosen the best response.7I just happened to discover this question in Math Stack Exchange (MSE): http://math.stackexchange.com/questions/54506/isthisbatmanequationforreal/54568#54568 Here's the image: dw:1435886055793:dw http://i.stack.imgur.com/VYKfg.jpg I tried to graph it on Desmos: https://www.desmos.com/calculator/cscx2zcrlf Graphing each factor separately turns out well, however when I multiply all factors, graph won't appear. I think Desmos just choke on it, so I tried something simple; https://www.desmos.com/calculator/enxuzekis6 Yet it doesn't show anything... So apparently image above is not true. My questions are: \(\Large 1)\) Is there any way to union relations? By "union," I mean in \(\mathbb R^2\), if you union \(f(x,y)=0\) and \(g(x,y)=0\), then you would have new union relation \(U(x,y)=0\), which if you graph it, then \(f(x,y)=0\) and \(g(x,y)=0\) would both be graphed simultaneously. From image, it's \(U(x,y) = f(x,y)g(x,y)=0\), but it seems that it is not true. \(\Large 2)\) Why does no one seem to notice that image is not true in MSE question?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A relation is a set, so obviously you could union two relations.

geerky42
 one year ago
Best ResponseYou've already chosen the best response.7Yeah I know. Not sure what's correct word for that, but let's just stick with "union" By "union," I mean saying if you graph \(f(x,y)=0\) and \(g(x,y)=0\), you would see something on xy plane. If we "union" these relations, then we would have new relation \(U(x,y)=0\), such that if you graph \(U(x,y)\), then you would see same thing as if you graph \(f(x,y)=0\) and \(g(x,y)=0\) separately at the same time.

geerky42
 one year ago
Best ResponseYou've already chosen the best response.7From shown image, you just multiply relations, but to me, it doesn't work.

just_one_last_goodbye
 one year ago
Best ResponseYou've already chosen the best response.0In my opinion it doesn't work :/ very complex and in testing ur not going to really make a perfect batman sign on the graph xD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0We are such math nerds XD

geerky42
 one year ago
Best ResponseYou've already chosen the best response.7For example; saying we have \(f(x,y) = x^2+y^2 1= 0\) and \(g(x,y) = (x+2)^2+y^21=0\) Then \(U(x,y)\) would graph this:dw:1435887037303:dw Exactly what is \(U(x,y)\)? and how can I "combine" f(x,y) and g(x,y) to form U(x,y)?

geerky42
 one year ago
Best ResponseYou've already chosen the best response.7Wut now it works. https://www.desmos.com/calculator/uezz1quzdy Now I am confused lol...

geerky42
 one year ago
Best ResponseYou've already chosen the best response.7Got something to do with restrictions, I guess. That circles equation above here has no restriction, but Batman Equation does.

geerky42
 one year ago
Best ResponseYou've already chosen the best response.7https://www.desmos.com/calculator/enxuzekis6 First relation is restricted to \(x>1\) Second relation is restricted to \(x<1\) So together they would not appear in graph simply because they would be somewhere in complex. Just like Batman Equation. Is there any way to prevent that? OK I think that's too much to ask lol...

geerky42
 one year ago
Best ResponseYou've already chosen the best response.7@iambatman ( ͡° ͜ʖ ͡°)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im batman (in husky batman voice) o.o

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, the roots of \(f\cdot g\) is the union of the roots of \(g\) and the roots of \(f\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Only exceptions are when you have different domains

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You intersect the domains, then you union the roots in that intersected domain

geerky42
 one year ago
Best ResponseYou've already chosen the best response.7OK, I think I asked rather silly and pointless question. But thanks for put in effort for me.

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.0nincompoop I am afraid you made a mistake, it should be: Superman < Batman

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0That's super cool. I wonder if there will be a bunch of equations that can produce the Avengers Logo? When I was taking Calculus III there were some equations in polar coordinates that produced flowers, the number 8, or the infinity symbol.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0Anyway, graphing these equations separately was a good idea. If it was bunched up together, wouldn't something overlap and eventually the Batman Logo will no longer appear?

TheSmartOne
 one year ago
Best ResponseYou've already chosen the best response.0@iambatman should know xD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Haha, I believe it was here I posted something similar long time ago, in any case it seems you've figured out what your problem was? @geerky42 @nincompoop I resent that!

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.0FYI  Here is the Batman Equation implemented in Geogebra: \( \href{http:///www.geogebra.org/m/114}{Batman}\)
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